### Heat Transfer Area by NTU Method

Hot gases enter a finned tube heat exchanger at 300oC and leave at 100oC. It is used to heat water at a flow rate of 1 kg/s from 35oC to 125oC. The specific heat of exhaust hot gas is 1000 J/kg.K and the overall heat transfer coefficient based on the gas side is Uh = 100 W/m2.K. Determine the required heat transfer surface area using the NTU method

Calculations:

Since flow configuration is not given, we shall take it as countercurrent flow.

Mass flow rate of hot gases (mg) is obtained by energy balance:

1 x 4184 x (125 - 35) = mg x 1000 x (300 - 100)

mg = 1.8828 kg/s

Cmin = 1.8828 x 1000 = 1882.8 W/oC = Ch

Cmax = 1 x 4184 = 4184 W/oC = Cc

ε = Cc(Tc,out - Tc,in) / [Cmin(Th,in - Tc,in)] = 4184 x (125 - 35) / [1882.8 x (300 - 35)]

= 0.755

C = Cmin/Cmax = 1882.8/4184 = 0.45

For the counter flow exchanger, effectiveness - NTU relationship is given by,

ε = {1 - exp[-N(1 -C)]} / {1 - Cexp[-N(1 - C)]}

We know ε and C; and we have to find N. This has to be solved by iteration.

0.755 = {1 - exp[-N(1 - 0.45)]} / {1 - 0.45 exp[-N(1 - 0.45)]}

0.755 = (1 - e-0.55N) / (1 - 0.45e-0.55N)

starting with an initial assumption of N = 2,

(1 - e-0.55N) / (1 - 0.45e-0.55N) = 0.667 / 0.85 = 0.7845

by assuming N = 1.9,

(1 - e-0.55N) / (1 - 0.45e-0.55N) = 0.648 / 0.842 = 0.77

by assuming N = 1.8,

(1 - e-0.55N) / (1 - 0.45e-0.55N) = 0.628 / 0.833 = 0.754

Since the assumption of N = 1.8, almost balances the equation, we shall take N = 1.8 as the correct value.

The required gas surface area, A = NTU Cmin / U = 1.8 x 1882.8 / 100 = 33.89 m2